In this piece of work using only three grid points, we propose two sets of numerical methods in a coupled manner for the solution of fourth-order ordinary differential equation

Consider the fourth-order boundary value problem

Fourth-order differential equations occur in a number of areas of applied mathematics, such as in beam theory, viscoelastic and inelastic flows, and electric circuits. Some of them describe certain phenomena related to the theory of elastic stability. A classical fourth-order equation arising in the beam-column theory is the following (see Timoshenko [

The existence and uniqueness of solutions of boundary value problems are discussed in the papers and book of Agarwal and Krishnamoorthy, Agarwal and Akrivis (see [

In this paper we present the finite difference methods for the solution of fourth-order boundary value problem. We discretize the interval

Using the second-order central differences, we obtain a five-point difference formula for (

Since we need to solve a coupled system of equations at each mesh point, the block successive overrelaxation (BSOR) iterative method is used.

The method is described as follows: for

Note that

For the derivation of the method we follow the approaches given by Chawla [

At the grid point

Using the Taylor expansion about the grid point

Now, we need the

Thus, we obtain

Next, we obtain the

Therefore,

Thus, we obtain

Let

From (

Also, from (

Thus, we have the following result.

Consider the fourth-order nonlinear ordinary differential equation (

If the differential (

The block successive over relaxation (BSOR) method (See Mohanty and Evans [

If

We first write the difference equations (

Let

be the Jacobian of

are the

In the NBSOR method starting with any initial approximation

Consider the model problem

Let us denote by

The system of (

The rate of convergence of the BSOR method is dependent on the eigenvalues of

Hence, we can determine the optimal parameter as

Thus, we can determine the convergence factor

Hence, we get the convergence.

Thus, we have the following result.

The iteration method of the form (

Now, we discuss the stability analysis.

An iterative method for (

The above iterative method in matrix form can be written as

The eigenvalues of

Thus the eigenvalues of

Consider a singular fourth-order linear ordinary differential equation of the form

The above equation represents fourth-order ordinary differential equation in cylindrical polar coordinates.

The boundary conditions are given by

Applying the difference scheme (

Note that the scheme fails when the solution is to be determined at

Using the approximation (

Similarly, using the difference scheme (

Finite difference equations (

Consider the coupled nonlinear singular equations

The second-order difference scheme for solving the system (

The fourth-order difference scheme for solving the system (

The scheme (

Consider the boundary value problem [

This arises from the time-dependent Navier-Strokes equations for axisymmetric flow of an incompressible fluid contained between infinite disks that occupy the planes

To illustrate our method and to demonstrate computationally its convergence, we have solved the following linear problem using the BGS method (See [^{−12}. All computations were performed using double length arithmetic.

The problem is to solve (

The RMSE.

2nd-order method | 4th-order method | ||||

MAE | RMSE | MAE | RMSE | ||

1/8 | 0.9512 (−03) | 0.6737 (−03) | 0.6018 (−03) | 0.4089 (−03) | |

0.3162 (−02) | 0.2137 (−02) | 0.2961 (−02) | 0.1402 (−02) | ||

1/16 | 0.2212 (−03) | 0.1501 (−03) | 0.3799 (−04) | 0.2502 (−04) | |

0.7021 (−03) | 0.4995 (−03) | 0.3022 (−03) | 0.9846 (−04) | ||

1/32 | 0.5410 (−04) | 0.3576 (−04) | 0.1989 (−05) | 0.1304 (−05) | |

0.1729 (−03) | 0.1206 (−03) | 0.2302 (−05) | 0.5629 (−05) | ||

1/64 | 0.1339 (−04) | 0.8769 (−05) | 0.9996 (−07) | 0.6578 (−07) | |

0.4298 (−04) | 0.2972 (−04) | 0.1562 (−05) | 0.2983 (−06) | ||

1/128 | 0.3363 (−05) | 0.2174 (−05) | 0.4198 (−08) | 0.3345 (−08) | |

0.1072 (−04) | 0.7383 (−05) | 0.9892 (−07) | 0.1538 (−07) |

The boundary value problem is to solve (

The MAE and RMSE.

2nd-order method | 4th-order method | ||||

MAE | RMSE | MAE | RMSE | ||

1/8 | 0.8435 (−04) | 0.5652 (−04) | 0.2744 (−05) | 0.1880 (−05) | |

0.3552 (−03) | 0.2322 (−03) | 0.1058 (−04) | 0.7178 (−05) | ||

1/16 | 0.2292 (−04) | 0.1499 (−04) | 0.1854 (−06) | 0.1217 (−06) | |

0.7762 (−04) | 0.5391 (−04) | 0.6695 (−06) | 0.4351 (−06) | ||

1/32 | 0.5844 (−05) | 0.3771 (−05) | 0.1151 (−07) | 0.7446 (−08) | |

0.1885 (−04) | 0.1322 (−04) | 0.4115 (−07) | 0.2618 (−07) | ||

1/8 | 0.1961 (−02) | 0.1401 (−02) | 0.5662 (−04) | 0.3877 (−04) | |

0.6514 (−02) | 0.4687 (−02) | 0.3824 (−03) | 0.1910 (−03) | ||

1/16 | 0.4808 (−03) | 0.3312 (−03) | 0.3821 (−05) | 0.2558 (−05) | |

0.1597 (−02) | 0.1110 (−02) | 0.1929 (−04) | 0.9922 (−05) | ||

1/32 | 0.1197 (−03) | 0.8103 (−04) | 0.2435 (−06) | 0.1609 (−06) | |

0.4023 (−03) | 0.2714 (−03) | 0.1135 (−05) | 0.5843 (−06) |

The system of nonlinear equation (

The MAE and RMSE.

2nd-order method | 4th-order method | ||||

MAE | RMSE | MAE | RMSE | ||

0.7141 (−05) | 0.4859 (−05) | 0.2673 (−05) | 0.1818 (−05) | ||

0.3339 (−05) | 0.2240 (−05) | 0.6974 (−06) | 0.4745 (−06) | ||

0.2488 (−04) | 0.1683 (−04) | 0.1266 (−04) | 0.6538 (−05) | ||

0.1126 (−04) | 0.8137 (−05) | 0.2548 (−05) | 0.1656 (−05) | ||

0.1806 (−05) | 0.1182 (−05) | 0.2467 (−06) | 0.1639 (−06) | ||

0.8550 (−06) | 0.5595 (−06) | 0.4846 (−07) | 0.3212 (−07) | ||

0.6486 (−05) | 0.4128 (−05) | 0.1167 (−05) | 0.5946 (−06) | ||

0.2659 (−05) | 0.1957 (−05) | 0.1795 (−06) | 0.1119 (−06) | ||

0.4514 (−06) | 0.2917 (−06) | 0.2048 (−07) | 0.1346 (−07) | ||

0.2152 (−06) | 0.1391 (−06) | 0.3315 (−08) | 0.2161 (−08) | ||

0.1620 (−05) | 0.1020 (−05) | 0.1027 (−06) | 0.4911 (−07) | ||

0.6757 (−06) | 0.4828 (−06) | 0.1222 (−07) | 0.7515 (−08) | ||

0.1125 (−06) | 0.7216 (−07) | 0.1510 (−08) | 0.9836 (−09) | ||

0.5307 (−07) | 0.3442 (−07) | 0.2931 (−09) | 0.1883 (−09) | ||

0.4038 (−06) | 0.2525 (−06) | 0.8190 (−08) | 0.3662 (−08) | ||

0.1686 (−06) | 0.1192 (−06) | 0.9925 (−09) | 0.6537 (−09) | ||

0.3050 (−07) | 0.1946 (−07) | 0.1213 (−09) | 0.7945 (−10) | ||

0.1461 (−07) | 0.9331 (−08) | 0.1358 (−10) | 0.8769 (−11) | ||

0.1076 (−06) | 0.6812 (−07) | 0.6420 (−09) | 0.2913 (−09) | ||

0.4591 (−07) | 0.3237 (−07) | 0.5009 (−10) | 0.3048 (−10) |

Observe that for the linear singular problem for

The numerical results confirm that the proposed finite difference methods yield second- and fourth-order convergence for the solution and its derivative for the fourth-order ordinary differential equation. Difference formulas for mesh points near the boundary are obtained without using the fictitious points. The proposed method is applicable to problems in polar coordinates and the derivative of the solution is obtained as the by-product of the method. We employ the BGS method to solve the block matrix systems of the linear singular problem and the BSOR method to solve the nonlinear singular problem. We have solved here a physical problem that arises in the axisymmetric flow of an incompressible fluid.

This research was supported by “The Council of Scientific and Industrial Research” under Research Grant no. 09/045(0836)2009-EMR-I. The authors thank the reviewers for their valuable suggestions, which substantially improved the standard of the paper.